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olasank [31]
3 years ago
14

under what conditions would a scale record a weight for an object which is equal to the magnitude of the gravitational force on

the object?
Physics
2 answers:
pentagon [3]3 years ago
7 0

answer:if there are no external forces acting on the object

Explanation:

under what conditions would a scale record a weight for an object which is equal to the magnitude of the gravitational force on the object?

If gravity and the normal reaction force are the only forces acting on the body, and the body is at rest or moving at constant velocity, the normal force and the weight are equal and opposite. Also if there are no influence of external forces

and air resistance is negligible.

from newtons law of motion which states that the rate of change in momentum is directly proportional to the force applied

f=km(v-u)/t

f=kma

where k is the constant and  equal to 1

f=ma

d1i1m1o1n [39]3 years ago
4 0
In a non accelerated frame, also called inertial frame, when no additional forces are acting upon the body.

For instance, a scale in free fall will not record the weight, but zero, like the astronauts. But also, if someone is pulling the object or pushing it, the scale will record some other value.
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How is the moment magnitude scale used to describe earthquakes?
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Read 2 more answers
If a car is moving to the left with constantvelocity, one can conclude that
jeka94

Answer:

The net force applied to the car is zero.

Explanation:

We are given that a car is moving to the left with constant velocity.

When the car moving with constant velocity

Then, the final velocity=Initial velocity

Change in velocity=Final velocity- initial velocity=0

When change in velocity is zero then , acceleration of car

a=\frac{change\;in\;velocity}{time}=\frac{0}{t}=0

When acceleration is zero then, By Newtons second law

Force=Mass\times acceleration=Mass\times 0=0

The net force applied on the car will be zero.

Option C:The net force applied to the car is zero.

5 0
3 years ago
A projectile is fired at an upward angle of 45.0º from the top of a 265-m cliff with a speed of .sm 185 What will be its speed w
nignag [31]

The final velocity of the projectile when it strikes the ground below is 198.51 m/s.

<h3>Time of motion of the projectile</h3>

The time taken for the projectile to fall to the ground is calculated as follows;

h = vt + ¹/₂gt²

where;

  • h is height of the cliff
  • v is velocity
  • t is time of motion

265 = (185 x sin45)t + (0.5)(9.8)t²

265 = 130.8t + 4.9t²

4.9t² + 130.8t - 265 = 0

solve the quadratic equation using formula method,

t = 1.89 s

<h3>Final velocity of the projectile</h3>

vyf = vyi + gt

where;

  • vyf is the final vertical velocity
  • vyi is initial vertical velocity

vyf = (185 x sin45) + (9.8 x 1.89)

vyf = 149.322 m/s

vxf = vxi

where;

  • vxf is the final horizontal velocity
  • vxi is the initial horizontal velocity

vxf = 185 x cos(45)

vxf = 130.8 m/s

vf = √(vyf² + vxf²)

where;

  • vf is the speed of the projectile when it strikes the ground below

vf = √(149.322²  +  130.8²)

vf = 198.51 m/s

Learn more about final velocity here: brainly.com/question/6504879

#SPJ1  

3 0
2 years ago
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